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Solid mensuration · Lesson 2 of 3

Cones and pyramids

Volume as one third of base × height, the slant height from Pythagoras, the curved surface πrl, and a cone made by bending a sector.

18 minYou should already know: Plane mensuration
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A pyramid rises from a flat base to a point (the vertex). A cone is a pyramid with a circular base. Both hold exactly one third as much as the prism with the same base and height:

volume=13×base area×height\text{volume} = \frac13 \times \text{base area} \times \text{height}
hbase
PyramidV=13×base×hV = \frac13 \times \text{base} \times h
hrl
ConeV=13πr2hV = \frac13\pi r^2 h

So a cone has volume 13πr2h\frac13\pi r^2 h. In both drawings the dashed height hh goes straight down from the top to the centre of the base, at right angles to it.

More: volumes of cones and pyramids

Heights and slant heights

Three lengths go together in a cone, and they make a right-angled triangle: the height hh (from the vertex straight down to the centre of the base), the base radius rr, and the slant height ll (from the vertex down the side to the edge of the base).

l2=r2+h2l^2 = r^2 + h^2
hrll² = r² + h²
Height, radius, slant heightA right-angled triangle inside every cone

For a right pyramid, the vertex is directly above the centre of the base, where the diagonals cross. The slant edge, half the base’s diagonal and the height make the right-angled triangle.

Worked example · WAEC 2014

WAEC 2014 · Paper 2 · Q5 (a, c)

The diagram shows a right pyramid with a rectangular base WXYZWXYZ and vertex OO. If ∣WX∣=8 cm|WX| = 8\text{ cm}, ∣ZW∣=6 cm|ZW| = 6\text{ cm} and ∣OX∣=13 cm|OX| = 13\text{ cm}, calculate the:

8 cm6 cm13 cmWXYZO
Not to scale.

height of the pyramid;

volume of the pyramid.

  1. Half the diagonal

    The diagonal ∣ZX∣=82+62=10|ZX| = \sqrt{8^2 + 6^2} = 10 cm, so from the centre MM to the corner XX is 5 cm.

    Think first. The base is 8 cm by 6 cm. How long is the diagonal?

  2. (a) The height

    ∣OM∣=132−52=144=12|OM| = \sqrt{13^2 - 5^2} = \sqrt{144} = 12 cm.

    Think first. Triangle OMXOMX has a right angle at MM and hypotenuse OX=13OX = 13.

  3. (c) The volume

    13×(8×6)×12=192 cm3\frac13 \times (8 \times 6) \times 12 = 192\text{ cm}^3

More: the height of a pyramid

The surface of a cone

curved surface=πrltotal, solid cone=πrl+πr2\begin{aligned} \text{curved surface} &= \pi r l \\ \text{total, solid cone} &= \pi r l + \pi r^2 \end{aligned}

More: the surface of a cone

A cone from a sector

Cut a sector out of a sheet and bend it until its straight edges meet: you get a cone.

Bend a sector into a coneChange the angle of the sector
135°sector: radius 28hrl = 28the cone
10.5base radius r = 135/360 × 2825.96height h = √(l² − r²)924curved surface πrl = sector area2998volume ⅓πr²h
Join the two straight edges. The sector's radius becomes the slant height l = 28. Its arc (66) goes round the base, so 2πr = 135/360 × 2πl, which gives r = 135/360 × 28 = 10.5. Then h = √(28² − 10.5²) = 25.96. The vertical angle at the top is 2 × sin⁻¹(r/l) ≈ 44°.

As you bend it:

  • the sector’s radius becomes the cone’s slant height ll;
  • the sector’s arc becomes the circumference of the base, so 2πr=θ360×2πl2\pi r = \frac{\theta}{360} \times 2\pi l, which gives r=θ360×lr = \frac{\theta}{360} \times l;
  • the sector’s area becomes the cone’s curved surface.
θllarc → base circler = θ ÷ 360 × l
Bending a sectorThe radius becomes the slant height; the arc becomes the base circle

More: a cone from a sector

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q8 (b)

A sector of angle 220∘220^\circ is removed from a thin circular metal sheet of radius 63 cm63\text{ cm}. It is then folded with the straight edges meeting to form a right circular cone. Calculate, correct to one decimal place, the: (i) base radius; (ii) volume, of the cone. [Take π=227]\left[\text{Take }\pi = \frac{22}{7}\right]

  1. (i) The base radius

    The arc becomes the base’s circumference: r=220360×63=38.5r = \frac{220}{360} \times 63 = 38.5 cm.

    Think first. What does the sector's arc become?

  2. The height

    The slant height is the sheet’s radius, 63 cm: h=632−38.52=2486.75≈49.867h = \sqrt{63^2 - 38.5^2} = \sqrt{2486.75} \approx 49.867 cm.

    Think first. What is the slant height? Then use Pythagoras.

  3. (ii) The volume

    13×227×38.52×49.867≈77 435.6 cm3\begin{aligned} &\tfrac13 \times \tfrac{22}{7} \times 38.5^2 \times 49.867 \\ &\approx 77\,435.6\text{ cm}^3 \end{aligned}

Your turn

WAEC 2021 · Paper 2 · Q3

The curved surface area of a cone is 242 cm2242\text{ cm}^2. If the slant height is 4 cm4\text{ cm} more than the radius, calculate, correct to one decimal place, the: [Take π=227]\left[\text{Take }\pi = \frac{22}{7}\right]

  1. (a)

    radius;

  2. (b)

    height;

  3. (c)

    volume of the cone.

Worked solution (try it first)

(a)

  1. Let the radius be rr.
  2. Then l=r+4l = r + 4.
  3. Curved surface: 227×r(r+4)=242\frac{22}{7} \times r(r + 4) = 242, so r2+4r=77r^2 + 4r = 77 and r2+4r−77=0r^2 + 4r - 77 = 0.
  4. Factorise: (r+11)(r−7)=0(r + 11)(r - 7) = 0.
  5. The radius is positive, so r=7.0r = 7.0 cm.

(b)

  1. l=11l = 11 cm, so h=112−72h = \sqrt{11^2 - 7^2}
    =72= \sqrt{72}
    ≈8.5\approx 8.5 cm.

(c)

  1. Volume =13×227×72×72= \frac13 \times \frac{22}{7} \times 7^2 \times \sqrt{72}
    ≈435.6 cm3\approx 435.6\text{ cm}^3.

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